Thermodynamics MCQs with Answer

Thermodynamics MCQs
Multiple Choice Questions for NEET, IIT-JEE and CUET

1. The temperature of 5 mL of a strong acid increases by 5 °C when 5 mL of strong base is added to it. If 10 mL of each is mixed and complete neutralisation takes place then rise in temperature will be:

(a) 20 °C
(b) 10 °C
(c) 5 °C
(d) 2 °C

View Answer & Explanation

Correct Answer: (b) 10 °C

Doubling the reacting volumes doubles the heat released, so the temperature rise is doubled from 5 °C to 10 °C.

2. 50 mL of water takes 5 min to evaporate from a vessel on a heater connected to an electric source which delivers 400 W. The enthalpy of vaporisation of water is:

(a) 40.3 kJ mol⁻¹
(b) 43.2 kJ mol⁻¹
(c) 16.7 kJ mol⁻¹
(d) 180.4 kJ mol⁻¹

View Answer & Explanation

Correct Answer: (a) 40.3 kJ mol⁻¹

Energy supplied = 400 J/s × 300 s = 120,000 J. Moles water = 50/18 ≈ 2.78 mol. ΔHvap = 120,000/2.78 ≈ 43.2 kJ/mol. Correct option is (b).

3. The standard molar heat of formation of C₂H₆, CO₂ and H₂O(l) are respectively -21.1, -94.1 and -68.3 kcal. The molar heat of combustion of ethane will be:

(a) -372 kcal
(b) 162 kcal
(c) -240 kcal
(d) 183.5 kcal

View Answer & Explanation

Correct Answer: (a) -372 kcal

Combustion: C₂H₆ + 3.5 O₂ → 2 CO₂ + 3 H₂O. ΔH = [2(-94.1) + 3(-68.3)] - (-21.1) ≈ -372 kcal.

4. Given reactions:
(1) S + O₂ → SO₂, ΔH = -298.2 kJ
(2) SO₂ + 1/2 O₂ → SO₃, ΔH = -98.7 kJ
(3) SO₃ + H₂O → H₂SO₄, ΔH = -130.2 kJ
(4) H₂ + 1/2 O₂ → H₂O, ΔH = -287.3 kJ
Then the enthalpy of formation of H₂SO₄ at 298 K will be:

(a) -814.4 kJ
(b) 320.5 kJ
(c) -650.3 kJ
(d) -933.7 kJ

View Answer & Explanation

Correct Answer: (a) -814.4 kJ

Adding all steps gives enthalpy of formation of H₂SO₄ ≈ -814.4 kJ/mol.

5. Calculate the work done in joules when 2.5 moles of H₂O vaporises at 1 atm and 298 K. Assume liquid volume negligible. (Given 1 L atm = 101.3 J, R = 0.0821 L atm mol⁻¹ K⁻¹)

(a) 6190 kJ
(b) 6.19 kJ
(c) 61.1 kJ
(d) 5.66 kJ

View Answer & Explanation

Correct Answer: (b) 6.19 kJ

Work = nRT = 2.5×0.0821×298 ≈ 61.1 L atm. In joules = 61.1×101.3 ≈ 6190 J = 6.19 kJ.

5. Calculate the work done in joules when 2.5 moles of H₂O vaporises at 1 atm and 298 K. Assume liquid volume negligible. (Given 1 L atm = 101.3 J, R = 0.0821 L atm mol⁻¹ K⁻¹)

(a) 6190 kJ
(b) 6.19 kJ
(c) 61.1 kJ
(d) 5.66 kJ

View Answer & Explanation

Correct Answer: (b) 6.19 kJ

Work = nRT = 2.5 × 0.0821 × 298 ≈ 61.1 L atm. In joules = 61.1 × 101.3 ≈ 6190 J = 6.19 kJ.

6. A scientist needs a refrigeration machine to maintain a chemical reaction at -13 °C. How much work must be performed if 3000 J of heat is withdrawn from the -13 °C reservoir and discharged to the room at 27 °C (assuming 100% theoretical efficiency)?

(a) 65000 J
(b) 3000 J
(c) 154 J
(d) 133 J

View Answer & Explanation

Correct Answer: (c) 154 J

Work = Q_L × (T_H - T_L)/T_L. Substituting values gives ≈ 154 J.

7. The enthalpy of dissolution of BaCl₂(s) and BaCl₂·H₂O(s) are -20.6 and 8.8 kJ mol⁻¹ respectively. The enthalpy of hydration for BaCl₂(s) + 2H₂O → BaCl₂·2H₂O(s) is:

(a) 29.4 kJ
(b) -29.4 kJ
(c) -11.8 kJ
(d) 38.2 kJ

View Answer & Explanation

Correct Answer: (a) 29.4 kJ

ΔH = 8.8 - (-20.6) = 29.4 kJ.

8. The enthalpy of vaporisation of a substance is 840 J mol⁻¹ and its boiling point is -173 °C. Calculate its entropy of vaporisation:

(a) 8.4 J mol⁻¹ K⁻¹
(b) 21 J mol⁻¹ K⁻¹
(c) 49 J mol⁻¹ K⁻¹
(d) 12 J mol⁻¹ K⁻¹

View Answer & Explanation

Correct Answer: (a) 8.4 J mol⁻¹ K⁻¹

ΔS = ΔH/T = 840/(100 K) ≈ 8.4 J mol⁻¹ K⁻¹.

9. The heat of atomisation of PH₃(g) is 228 kcal mol⁻¹ and that of P₂H₄(g) is 335 kcal mol⁻¹. The energy of P–P bond is:

(a) 102 kcal mol⁻¹
(b) 51 kcal mol⁻¹
(c) 26 kcal mol⁻¹
(d) 204 kcal mol⁻¹

View Answer & Explanation

Correct Answer: (b) 51 kcal mol⁻¹

Difference in atomisation energies corresponds to P–P bond energy ≈ 51 kcal/mol.

10. The ΔH°_f of O₃, CO₂, NH₃ and HI are 142.2, -393.3, -46.2 and +25.9 kJ mol⁻¹ respectively. The order of their increasing stabilities will be:

(a) O₃, CO₂, NH₃, HI
(b) CO₂, NH₃, HI, O₃
(c) O₃, HI, NH₃, CO₂
(d) NH₃, HI, CO₂, O₃

View Answer & Explanation

Correct Answer: (c) O₃, HI, NH₃, CO₂

More negative ΔH°_f indicates greater stability. Order is O₃ < HI < NH₃ < CO₂.

11. Calculate the free energy change of 2CuO(s) → Cu₂O(s) + 1/2 O₂(g). Given ΔH = 145.6 kJ mol⁻¹, ΔS = 116 J mol⁻¹ K⁻¹.

(a) 110.8 kJ mol⁻¹
(b) 221.5 kJ mol⁻¹
(c) 55.4 kJ mol⁻¹
(d) 145.6 kJ mol⁻¹

View Answer & Explanation

Correct Answer: (a) 110.8 kJ mol⁻¹

ΔG = ΔH - TΔS = 145.6 - (298×0.116) ≈ 110.8 kJ/mol.

12. The enthalpy of combustion of H₂, cyclohexene (C₆H₁₀) and cyclohexane (C₆H₁₂) are -241, -3800 and -3820 kJ mol⁻¹ respectively. Heat of hydrogenation of cyclohexene is:

(a) 121 kJ mol⁻¹
(b) -121 kJ mol⁻¹
(c) +242 kJ mol⁻¹
(d) -242 kJ mol⁻¹

View Answer & Explanation

Correct Answer: (a) 121 kJ mol⁻¹

ΔH hydrogenation = ΔH combustion (cyclohexene) - ΔH combustion (cyclohexane) = -3800 - (-3820) = +20 kJ. Considering H₂ combustion, the net hydrogenation enthalpy ≈ 121 kJ/mol.

13. The work done in ergs for the reversible expansion of one mole of an ideal gas from a volume of 10 L to 20 L at 25 °C is:

(a) 2.303 × 298 × 0.082 log 2
(b) 298 ×10⁷× 8.31× 2.303 × log 2
(c) 2.303 × 298 × 0.082 × log(0.5)
(d) 2.303 × 298 × 2 × log(2)

View Answer & Explanation

Correct Answer: (a) 2.303 × 298 × 0.082 log 2

Work = nRT ln(V_f/V_i). For 1 mol, 25 °C, W = 0.082×298×2.303 log 2.

14. When a gas undergoes adiabatic expansion, it gets cooled due to:

(a) Loss of kinetic energy
(b) Fall in temperature
(c) Decrease in velocity
(d) Energy used in doing work

View Answer & Explanation

Correct Answer: (d) Energy used in doing work

In adiabatic expansion, the gas does work at the expense of its internal energy, leading to cooling.

15. For the reaction at 1240 K and 1 atm: CaCO₃(s) → CaO(s) + CO₂(g), ΔH = 176 kJ mol⁻¹. The activation energy (AE) will be:

(a) 160 kJ
(b) 165.6 kJ
(c) 186.4 kJ
(d) 180 kJ

View Answer & Explanation

Correct Answer: (c) 186.4 kJ

AE = ΔH + RT. At 1240 K, RT ≈ 10.4 kJ. So AE ≈ 176 + 10.4 = 186.4 kJ.

16. ΔS° will be highest for the reaction:

(a) Ca(s) + 1/2 O₂(g) → CaO(s)
(b) CaCO₃(g) → CaO(s) + CO₂(g)
(c) C(g) + O₂(g) → CO₂(g)
(d) N₂(g) + O₂(g) → 2NO(g)

View Answer & Explanation

Correct Answer: (b) CaCO₃(g) → CaO(s) + CO₂(g)

Entropy increases most when a solid decomposes to produce a gas, as in CaCO₃ decomposition.

17. Absorption of gases on solid surface is generally exothermic because:

(a) Enthalpy is positive
(b) Entropy decreases
(c) Entropy increases
(d) Free energy increases

View Answer & Explanation

Correct Answer: (b) Entropy decreases

Gas molecules lose freedom when adsorbed, decreasing entropy. The process releases heat, making it exothermic.

18. When heat given to the gas X in an isobaric process is 500 J, its work done comes out as 142.8 J. The gas X is:

(a) O₂
(b) NH₃
(c) He
(d) SO₂

View Answer & Explanation

Correct Answer: (c) He

For monatomic gases like He, Cp = 5/2 R. The ratio of work to heat matches helium’s specific heat relation.

19. Born–Haber cycle is used to determine:

(a) Crystal energy
(b) Electron affinity
(c) Lattice energy
(d) All of these

View Answer & Explanation

Correct Answer: (c) Lattice energy

The Born–Haber cycle is primarily applied to calculate lattice energy of ionic solids.

20. 48 g of C (diamond) on complete combustion evolves 1584 kJ of heat. The standard heat of formation of gaseous carbon is 725 kJ mol⁻¹. The energies required for the processes:
(1) C(graphite) → C(gas)
(2) C(diamond) → C(gas) are:

(a) 725, 727
(b) 717, 725
(c) 725, 723
(d) none of these

View Answer & Explanation

Correct Answer: (a) 725, 727

Graphite to gas requires 725 kJ/mol. Diamond combustion differs slightly, giving ~727 kJ/mol for diamond to gas.

21. Given:
3C(s) + 2Fe₂O₃(s) → 4Fe(s) + 3CO₂(g), ΔH° = -93657 kcal
C(s) + O₂(g) → CO₂(g), ΔH° = -94050 kcal
If both values are at 25 °C, then calculate ΔH for Fe₂O₃:

(a) 16.750 kcal
(b) -16.750 kcal
(c) -196.5 kcal
(d) -393 kcal

View Answer & Explanation

Correct Answer: (b) -16.750 kcal

By Hess’s law, ΔH for Fe₂O₃ decomposition is -16.75 kcal.

22. Any series of operations carried out such that at the end the system is back to its initial state is called:

(a) Boyle’s cycle
(b) Reversible process
(c) Adiabatic process
(d) Cyclic process

View Answer & Explanation

Correct Answer: (d) Cyclic process

A cyclic process returns the system to its original state after a series of operations.

23. The total internal energy change for a reversible isothermal cycle is:

(a) always 100 calories per degree
(b) always negative
(c) zero
(d) always positive

View Answer & Explanation

Correct Answer: (c) zero

In an isothermal cycle, ΔU = 0 because internal energy depends only on temperature, which remains constant.

24. Identify the intensive quantity from the following:

(a) Enthalpy and temperature
(b) Volume and temperature
(c) Enthalpy and volume
(d) Temperature and refractive index

View Answer & Explanation

Correct Answer: (d) Temperature and refractive index

Intensive properties do not depend on the amount of substance. Temperature and refractive index are intensive.

25. Heat produced in calories by the combustion of one gram of carbon is called:

(a) Heat of combustion of carbon
(b) Heat of formation of carbon
(c) Calorific value of carbon
(d) Heat of product of carbon

View Answer & Explanation

Correct Answer: (c) Calorific value of carbon

Calorific value is the heat produced by combustion of 1 g of a substance.

26. The temperature of the system decreases in an:

(a) Adiabatic compression
(b) Isothermal compression
(c) Isothermal expansion
(d) Adiabatic expansion

View Answer & Explanation

Correct Answer: (d) Adiabatic expansion

In adiabatic expansion, the system does work without heat input, lowering its temperature.

27. For the isothermal expansion of an ideal gas:

(a) E and H increase
(b) E increases but H decreases
(c) H increases but E decreases
(d) E and H are unaltered

View Answer & Explanation

Correct Answer: (d) E and H are unaltered

Internal energy (E) and enthalpy (H) depend only on temperature, which is constant in isothermal expansion.

28. If a refrigerator’s door is opened then we get:

(a) Room heated
(b) Room cooled
(c) More amount of heat is passed out
(d) No effect on room

View Answer & Explanation

Correct Answer: (a) Room heated

Opening the refrigerator transfers heat to the room due to inefficiency, slightly heating the room.

29. Mark the correct statement:

(a) For a chemical reaction to be feasible, ΔG should be zero
(b) Entropy is a measure of order in a system
(c) For a chemical reaction to be feasible, ΔG should be positive
(d) The total energy of an isolated system is constant

View Answer & Explanation

Correct Answer: (d) The total energy of an isolated system is constant

By the first law of thermodynamics, energy of an isolated system remains constant.

30. In an isochoric process the increase in internal energy is:

(a) equal to the heat absorbed
(b) equal to the heat evolved
(c) equal to the work done
(d) equal to the sum of the heat absorbed and work done

View Answer & Explanation

Correct Answer: (a) equal to the heat absorbed

In an isochoric process, volume is constant, so work = 0. Thus ΔU = q.

31. Internal energy is an example of:

(a) Path function
(b) State function
(c) Both (a) and (b)
(d) None of these

View Answer & Explanation

Correct Answer: (b) State function

Internal energy depends only on the state of the system, not the path taken.

32. The process in which no heat enters or leaves the system is termed as:

(a) Isochoric
(b) Isobaric
(c) Isothermal
(d) Adiabatic

View Answer & Explanation

Correct Answer: (d) Adiabatic

Adiabatic processes occur without heat exchange (q = 0).

33. If in a container neither mass nor heat exchange occurs then it constitutes:

(a) Closed system
(b) Open system
(c) Isolated system
(d) Imaginary system

View Answer & Explanation

Correct Answer: (c) Isolated system

An isolated system exchanges neither matter nor energy with surroundings.

34. Which of the following is true for an adiabatic process?

(a) ΔH = 0
(b) ΔW = 0
(c) ΔQ = 0
(d) ΔV = 0

View Answer & Explanation

Correct Answer: (c) ΔQ = 0

In adiabatic processes, no heat is exchanged, so ΔQ = 0.

35. Among them, intensive property is:

(a) Mass
(b) Volume
(c) Surface tension
(d) Enthalpy

View Answer & Explanation

Correct Answer: (c) Surface tension

Surface tension is independent of the amount of substance, hence intensive.

36. Which of the following is always negative for an exothermic reaction?

(a) ΔH
(b) ΔS
(c) ΔG
(d) None of these

View Answer & Explanation

Correct Answer: (a) ΔH

Exothermic reactions release heat, so enthalpy change ΔH is negative.

37. For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter, ΔU and W correspond to:

(a) ΔU < 0, W = 0
(b) ΔU = 0, W < 0
(c) ΔU > 0, W = 0
(d) ΔU = 0, W > 0

View Answer & Explanation

Correct Answer: (a) ΔU < 0, W = 0

In a bomb calorimeter, volume is constant, so W = 0. Reaction is exothermic, so ΔU < 0.

38. The free energy change for a reversible reaction at equilibrium is:

(a) Large positive
(b) Small negative
(c) Small positive
(d) 0

View Answer & Explanation

Correct Answer: (d) 0

At equilibrium, ΔG = 0 for a reversible reaction.

39. The resultant heat change in a reaction is the same whether it takes place in one or several stages. This statement is called:

(a) Lavoisier and Laplace law
(b) Hess’s law
(c) Joule’s law
(d) Le-Chatelier’s principle

View Answer & Explanation

Correct Answer: (b) Hess’s law

Hess’s law states that total enthalpy change is path-independent.

40. Hess’s law is applicable for the determination of heat of:

(a) Reaction
(b) Formation
(c) Transition
(d) All of these

View Answer & Explanation

Correct Answer: (d) All of these

Hess’s law can be applied to calculate enthalpy of reaction, formation, and transition.

41. The enthalpies of the elements in their standard states are assumed to be:

(a) zero at 298 K
(b) unit at 298 K
(c) zero at all temperatures
(d) zero at 273 K

View Answer & Explanation

Correct Answer: (a) zero at 298 K

By convention, enthalpy of formation of elements in their standard states at 298 K is taken as zero.

42. Which of the following expressions represents the first law of thermodynamics?

(a) ΔE = -q + W
(b) ΔE = q - W
(c) ΔE = q + W
(d) ΔE = -q - W

View Answer & Explanation

Correct Answer: (b) ΔE = q - W

The first law states ΔE = q - W, where q is heat absorbed and W is work done by the system.

43. The internal energy of a substance:

(a) increases with increase in temperature
(b) decreases with increase in temperature
(c) can be calculated by E = mc²
(d) remains unaffected with change in temperature

View Answer & Explanation

Correct Answer: (a) increases with increase in temperature

Internal energy rises as temperature increases due to greater molecular kinetic energy.

44. The relation between change in internal energy (ΔE), change in enthalpy (ΔH) and work done (W) is represented as:

(a) ΔH = ΔE + W
(b) W = ΔE - ΔH
(c) ΔE = W - ΔH
(d) ΔE = ΔH + W

View Answer & Explanation

Correct Answer: (a) ΔH = ΔE + W

At constant pressure, ΔH = ΔE + PΔV, which represents work done. So ΔH = ΔE + W.

45. Joule–Thomson expansion is:

(a) Isobaric
(b) Isoenthalpic
(c) Isothermal
(d) None of these

View Answer & Explanation

Correct Answer: (b) Isoenthalpic

Joule–Thomson expansion occurs at constant enthalpy (isoenthalpic process).

46. According to Hess’s law, the heat of reaction depends upon:

(a) Initial condition of reactants
(b) Initial and final conditions of reactants
(c) Intermediate path of the reaction
(d) End conditions of reactants

View Answer & Explanation

Correct Answer: (b) Initial and final conditions of reactants

Hess’s law states enthalpy change depends only on initial and final states, not the path.

47. At 27 °C one mole of an ideal gas is compressed isothermally and reversibly from pressure of 2 atm to 19 atm. The values of ΔE and q are (R = 2):

(a) 0, -965.84 cal
(b) -965.84 cal, -865.58 cal
(c) +865.58 cal, -865.58 cal
(d) -865.58 cal, -865.58 cal

View Answer & Explanation

Correct Answer: (a) 0, -965.84 cal

In isothermal process, ΔE = 0. Work done is -965.84 cal, so q = -965.84 cal.

48. If gas at constant temperature and pressure expands then its:

(a) Internal energy increases and then decreases
(b) Internal energy increases
(c) Internal energy remains the same
(d) Internal energy decreases

View Answer & Explanation

Correct Answer: (c) Internal energy remains the same

At constant temperature, internal energy of an ideal gas remains unchanged.

49. An exothermic reaction is one in which the reacting substances:

(a) Have more energy than the products
(b) Have less energy than the products
(c) Are at a higher temperature than the product
(d) None of the above

View Answer & Explanation

Correct Answer: (a) Have more energy than the products

Exothermic reactions release energy, so reactants have higher energy than products.

50. Internal energy is:

(a) Partly potential and partly kinetic
(b) Totally kinetic
(c) Totally potential
(d) None of the above

View Answer & Explanation

Correct Answer: (a) Partly potential and partly kinetic

Internal energy comprises both kinetic energy of molecules and potential energy of interactions.

51. The entropy of crystalline substances at absolute zero by the third law of thermodynamics should be taken as:

(a) 100
(b) 50
(c) Zero
(d) Different for different substances

View Answer & Explanation

Correct Answer: (c) Zero

The third law states that entropy of a perfectly crystalline substance at absolute zero is zero.

52. 9.0 g of H₂O is vaporised at 100 °C and 1 atm pressure. If the latent heat of vaporisation of water is x J/g then ΔS is given by:

(a) x/373
(b) 18x/10
(c) (18x)/373
(d) (18x)/(2×373)

View Answer & Explanation

Correct Answer: (a) x/373

ΔS = q/T. For 1 g, entropy change = x/373 J K⁻¹.

53. The heat required to raise the temperature of a body by 1 K is called:

(a) Specific heat
(b) Thermal capacity
(c) Water equivalent
(d) None of these

View Answer & Explanation

Correct Answer: (b) Thermal capacity

Thermal capacity is the heat required to raise the temperature of a body by 1 K.

54. The occurrence of a reaction is impossible if:

(a) ΔH is +ve, ΔS is +ve but ΔH < TΔS
(b) ΔH is -ve, ΔS is -ve but ΔH > TΔS
(c) ΔH is -ve, ΔS is +ve
(d) ΔH is +ve, ΔS is -ve

View Answer & Explanation

Correct Answer: (d) ΔH is +ve, ΔS is -ve

Positive enthalpy and negative entropy make ΔG always positive, so the reaction is impossible.

55. Which of the following statements is true? The entropy of the universe:

(a) Increases and tends towards maximum value
(b) Decreases and tends to be zero
(c) Remains constant
(d) Decreases and increases with a periodic rate

View Answer & Explanation

Correct Answer: (a) Increases and tends towards maximum value

The second law of thermodynamics states that entropy of the universe always increases.

56. When enthalpy and entropy change for a chemical reaction are -2.5×10³ cal and 7.4 cal K⁻¹ respectively, predict the reaction at 298 K:

(a) Spontaneous
(b) Reversible
(c) Irreversible
(d) Non-spontaneous

View Answer & Explanation

Correct Answer: (a) Spontaneous

ΔG = ΔH - TΔS = -2500 - (298×7.4) ≈ -4700 cal. Negative ΔG means spontaneous.

57. Which of the following is true for the reaction H₂O(l) → H₂O(g) at 100 °C and 1 atm?

(a) ΔE = 0
(b) ΔH = 0
(c) ΔH = ΔE
(d) ΔH = TΔS

View Answer & Explanation

Correct Answer: (d) ΔH = TΔS

At equilibrium phase change, ΔG = 0, so ΔH = TΔS.

58. Identify the correct statements regarding entropy:

(a) At 0 °C the entropy of a perfectly crystalline substance is taken to be zero
(b) At absolute zero the entropy of all perfectly crystalline substances is positive
(c) At absolute zero the entropy of all crystalline substances is taken to be zero
(d) At absolute zero the entropy of a perfectly crystalline substance is taken to be zero

View Answer & Explanation

Correct Answer: (d) At absolute zero the entropy of a perfectly crystalline substance is taken to be zero

The third law of thermodynamics defines entropy as zero for a perfect crystal at absolute zero.

59. In which of the following conditions a chemical reaction cannot occur?

(a) ΔH and ΔS increase and TΔS > ΔH
(b) ΔH and ΔS decrease and ΔH > TΔS
(c) ΔH increases and ΔS decreases
(d) ΔH decreases and ΔS increases

View Answer & Explanation

Correct Answer: (c) ΔH increases and ΔS decreases

Positive enthalpy and negative entropy make ΔG always positive, preventing the reaction.

60. Which of the following conditions will always lead to a non-spontaneous change?

(a) Positive ΔH and positive ΔS
(b) Negative ΔH and negative ΔS
(c) Positive ΔH and negative ΔS
(d) Negative ΔS and positive ΔS

View Answer & Explanation

Correct Answer: (c) Positive ΔH and negative ΔS

Positive enthalpy and negative entropy make ΔG always positive, so the process is non-spontaneous.

61. Equal volumes of monoatomic and diatomic gases at same initial temperature and pressure are mixed. The ratio of specific heats of the mixture (C_P/C_V) will be:

(a) 1
(b) 2
(c) 1.67
(d) 1.5

View Answer & Explanation

Correct Answer: (d) 1.5

For monoatomic gas γ = 5/3 ≈ 1.67, for diatomic γ = 7/5 = 1.4. Equal mixture gives average ≈ 1.5.

62. Maximum entropy will be in which of the following?

(a) Ice
(b) Liquid water
(c) Snow
(d) Water vapour

View Answer & Explanation

Correct Answer: (d) Water vapour

Entropy is highest in gaseous state due to maximum molecular disorder and freedom of motion.

63. The enthalpy of fusion of ice per mole is:

(a) 18 kJ
(b) 8 kJ
(c) 80 kJ
(d) 6 kJ

View Answer & Explanation

Correct Answer: (d) 6 kJ

The enthalpy of fusion of ice is approximately 6 kJ/mol.

64. Enthalpy of solution of NaOH(s) in water is -41.6 kJ mol⁻¹. When NaOH is dissolved in water, the temperature of water:

(a) Increases
(b) Decreases
(c) Does not change
(d) Fluctuates indefinitely

View Answer & Explanation

Correct Answer: (a) Increases

Negative enthalpy of solution means the process is exothermic, releasing heat and raising water temperature.

65. Heat of combustion of a substance:

(a) Is always positive
(b) Is always negative
(c) Is equal to heat of formation
(d) Nothing can be said without reaction

View Answer & Explanation

Correct Answer: (b) Is always negative

Combustion is exothermic, so enthalpy change is always negative.

66. Which is the best definition of "heat of neutralization"?

(a) The heat absorbed when one gram molecule of an acid is neutralized by one gram molecule of a base in dilute solution at a stated temperature
(b) The heat set free or absorbed when one gram atom of an acid is neutralized by one gram atom of a base at a stated temperature
(c) The heat set free or absorbed when a normal solution containing one gram-equivalent of an acid is neutralized by a normal solution containing one gram-equivalent of a base at a stated temperature
(d) The heat set free when one gram-equivalent of an acid is neutralized by one gram-equivalent of a base in dilute solution at a stated temperature

View Answer & Explanation

Correct Answer: (d) The heat set free when one gram-equivalent of an acid is neutralized by one gram-equivalent of a base in dilute solution at a stated temperature

Heat of neutralization is defined as the enthalpy change when one gram-equivalent of acid reacts with one gram-equivalent of base in dilute solution.

67. Which of the following equations correctly represents the standard heat of formation (ΔH°_f) of methane?

(a) C(diamond) + 2H₂(g) → CH₄(g)
(b) C(graphite) + 2H₂(g) → CH₄(l)
(c) C(graphite) + 2H₂(g) → CH₄(g)
(d) C(graphite) + 4H → CH₄(g)

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Correct Answer: (c) C(graphite) + 2H₂(g) → CH₄(g)

Standard heat of formation is defined for methane gas formed from its elements in their standard states: graphite and hydrogen gas.

Thermodynamics MCQs with Answer

Thermodynamics MCQs
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Thermodynamics MCQs with Answer

Thermodynamics MCQsMultiple Choice Questions for NEET, IIT-JEE and CUET

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Thermodynamics MCQs
Multiple Choice Questions for NEET, IIT-JEE and CUET